Question

How do you find the maclaurin series expansion of `sin(3x^2)`?

Answer 1

Use Maclaurin series for sin X. Substitute X = `3x^2`,
`sin(3x^2) = 3x^2-(3^3)x^6/(3!)+3^5x^10/(5!)-3^7x^14/(7!)\...`.

Explanation:

Maclaurin series for sin X:
`sin X = X-X^3/(3!)+X^5/(5!)-X^7/(7!)-...`.
So, `sin(3x^2) = 3x^2-(3^3)x^6/(3!)+3^5x^10/(5!)-3^7x^14/(7!)\...`
=. `sin(3x^2) = 3x^2-(3^3)x^6/(3!)+3^5x^10/(5!)-3^7x^14/(7!)\...`.